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3 years ago

Help im stuck on this problem

Mathematics

1 answer:

kotykmax [81]3 years ago

6 0

Answer:

${ \bf{m \angle A}} = { \bf{180 \degree -48 \degree }} \\ { \tt{m \angle A = 132 \degree}}$

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If your dog weighs 32 kilograms (kg), how many milligrams (mg) is this?

Tatiana [17]

Answer:

milligram would be 3.2e+7 or if not 3.2

7 0

2 years ago

Read 2 more answers

Lilly has read seven-eighths of her book and is on page 147. How many pages does lilly have left to read?

lara [203]

Answer:

Step-by-step explanation:

Since Lilly has read 7/8 of her book (aka 147 pages), divide 147 by 7 and you get 21.

This means 1/8=21

Since she has read 7/8 of the book, and the full book is 8/8, 8/8-7/8= 1/8

Meaning she has 1/8 of the book left to read.

As said before, 1/8 =21

Thus, she has 21 pages left to read.

4 0

2 years ago

What is the answer to this problem 18+4(28)

olganol [36]

Answer: 130

Step-by-step explanation: multiply 4 and 28 first because of PEMDAS. After you get 112, add 18 and you get 130.

5 0

3 years ago

Read 2 more answers

Haven't done this inna while just making sure I did it right

Viefleur [7K]

B is the correct answer
hope it helps!

7 0

2 years ago

The production department of Celltronics International wants to explore the relationship between the number of employees who ass

Flura [38]

Answer:

(a) Shown below

(b) There is a positive relation between the number of assemblers and production.

Step-by-step explanation:

Let X = number of assemblers and Y = number of units produced in an hour.

(a)

Consider the scatter plot below.

(b)

Based on the scatter plot it can be concluded that there is a positive relationship between the variables X and Y, i.e. as the value of X increases Y also increases.

(c)

The formula to compute the correlation coefficient is:

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }$

Compute the correlation coefficient between X and Y as follows:

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }=\frac{(5\times430)-(15\times120)}{\sqrt{[(5\times55)-15^{2}][(5\times3450)-120^{2}]}} =0.9272$

Thus, the correlation coefficient is 0.9272.

7 0

2 years ago